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Question: How much charge is required for the following reductions? (i) 1 mol of \(A{{l}^{3+}}\) to \(Al\) ...

How much charge is required for the following reductions?
(i) 1 mol of Al3+A{{l}^{3+}} to AlAl
(ii) 1 mol of Cu2+C{{u}^{2+}} to CuCu
(iii) 1 mol of MnO4−Mn{{O}_{4}}^{-} to Mn2+M{{n}^{2+}}

Explanation

Solution

Electrical energy is required for the reduction of unstable ions to their stable forms. Addition of electrons takes place in the electropositive ions to form stable compounds.

Complete step by step answer:
Let us primarily understand the basics of electrical energy and faraday's law.
Faraday-
It equals the amount of electrical charge that liberates one-gram equivalent of any ion from the electrolyte solution. It is the unit of electricity used in the study of electrochemical reactions.
It is a dimensionless unit of electrical charge, and equal to approximately 6.023×10236.023\times {{10}^{23}} electrical charge units (Avogadro’s constant).

Faraday’s law-
Now, in every electrochemical process, a certain amount of electrical charge is transferred during oxidation at one electrode and reduction at another. Thus, the law states that,
The amount of substance produced at each electrode is directly proportional to the quantity of charge flowing through the cell.
Key points-
The charge of one mole of electrons can be calculated by multiplying the charge of one electron by the Avogadro’s constant.
Thus, one mole of electron has a charge of 1 Faraday = 96500 C.
Illustration-
We have,
(i) 1 mol of Al3+A{{l}^{3+}} to AlAl
This can be stated as,
Al3++3e−→AlA{{l}^{3+}}+3{{e}^{-}}\to Al
The reduction will require 3 Faraday of electricity as 3 moles of electrons are involved.
Thus,
3×98500=2.895×105C3\times 98500=2.895\times {{10}^{5}}C

(ii) 1 mol of Cu2+C{{u}^{2+}} to CuCu
This can be stated as,
Cu2++2e−→CuC{{u}^{2+}}+2{{e}^{-}}\to Cu
The reduction will require 2 Faraday of electricity as 2 moles of electrons are involved.
Thus,
2×96500=1.93×105C2\times 96500=1.93\times {{10}^{5}}C

(iii) 1 mol of MnO4−Mn{{O}_{4}}^{-} to Mn2+M{{n}^{2+}}
This can be stated as,
MnO4−+5e−→Mn2+Mn{{O}_{4}}^{-}+5{{e}^{-}}\to M{{n}^{2+}}
The reduction will require 5 Faraday of electricity as 5 moles of electrons are involved.
Thus,
5×96500=4.825×105C5\times 96500=4.825\times {{10}^{5}}C

Note: 1 mole of electron has a charge of 1 faraday which is equal to 96500 Coulomb. Do not mix up anywhere.
Coulomb is the most commonly used unit for electricity but for solving word problems, conversion into Faraday units is available.